Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 16}{x + 10} = \dfrac{-18x - 96}{x + 10}$
Solution: Multiply both sides by $x + 10$ $ \dfrac{x^2 - 16}{x + 10} (x + 10) = \dfrac{-18x - 96}{x + 10} (x + 10)$ $ x^2 - 16 = -18x - 96$ Subtract $-18x - 96$ from both sides: $ x^2 - 16 - (-18x - 96) = -18x - 96 - (-18x - 96)$ $ x^2 - 16 + 18x + 96 = 0$ $ x^2 + 80 + 18x = 0$ Factor the expression: $ (x + 8)(x + 10) = 0$ Therefore $x = -8$ or $x = -10$ At $x = -10$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -10$, it is an extraneous solution.